Landau&Lifshitz's comment on WKB formula

Question

In L&L's quantum mechanics textbook $\S 46$, they derived the WKB approximated wave function at the classically forbidden region:

$$ \psi = \frac{C_1}{\sqrt{|p|}} \exp\left(-\frac{1}{\hbar} \int |p| \mathrm{d}x\right) + \frac{C_2}{\sqrt{|p|}} \exp\left(\frac{1}{\hbar} \int |p| \mathrm{d}x\right) \tag{46.10} $$

and they commented that:

It must, however, be borne in mind that the accuracy of the quasi-classical approximation is not such as to allow the retention in the wave function of exponentially small terms superimposed on exponentially large ones, and in this sense it is usually not permissible to retain both terms in (46.10).

I don't quite understand this comment, usually we ignore one term just because it is exponentially growing which is physically disallowed. But L&L seemed to argued that we must drop the exponentially small terms? I don't quite understand this comment, are there some examples?

And, in $\S 47$, they derived the connection condition for the WKB approximation. They seemed to use the fact:

If we follow the variation of the function (47.2) along the same semicircle in the opposite direction (from left to right), we see that at the beginning the first term rapidly becomes exponentially small in comparison with the second term. But the quasi-classical approximation does not allow us to include exponentially small terms in $\psi$ superimposed on the large principal term, and this is why the first term in (47.2) is "lost" in the passage along the semicircle.

Where equation (47.2) is the solution in the classical allowed region: $$ \psi = \frac{C_1}{\sqrt{p}}\exp\left(\frac{i}{\hbar}\int_a^x p \mathrm{d}x\right) + \frac{C_2}{\sqrt{p}}\exp\left(-\frac{i}{\hbar}\int_a^x p \mathrm{d}x\right) $$

But if we follow the circle in the complex plane described by L&L, at first the first term indeed exponentially decay, but after arriving the classical forbidden region, it becomes an exponentially growing term.

post related Landau & Lifshitz's Approach (contour method) on the WKB connection formulas, but it didn't answer my question.

Answer 1

Let's start with an easier example.

Say you have a function $f(x)$, and you expand it to quadratic order. Say you find $$ f(x) = x + x^2 + O(x^3) $$ Now say you want to add $g(x) = \frac{1}{2} x^3$ to $f(x)$. Should we say $f(x)+g(x)=x+x^2+\frac{1}{2} x^3 + \cdots$ ? No, because $f$ already has cubic order terms that we've neglected. By including the $\frac{1}{2} x^3$ term in $f+g$, we are implicitly claiming we "control" or "have calculated" the $x^3$ terms in both $f$ and $b$. But we have not done that. So the correct expression for $f+g$ to quadratic order is $$ f(x) + g(x) = x + x^2 + O(x^3) = f(x) $$ In other words, at this order, $g(x)$ does not contribute.

Something similar is happening in the WKB approximation. We have two solutions, like $$ f(x) = e^{p x}\left(1 + O(\frac{1}{p} ) \right) $$ and $$ g(x) = e^{-p x}\left((1 + O(\frac{1}{p} ) \right) $$ The leading order term in $g$, $\sim e^{-px}$, is smaller than $O(p^{-1} e^{p x})$ terms that we've already ignored in $f$. So if we evaluate $f+g$, then to the order we're working we should really think of $g$ as part of the error term, since we're already putting larger terms inside of the error term.